Question

A point object is kept in front of a convex spherical surface of radius of curvature $R$. Draw the ray diagram to show the formation of image and derive the relation between the object and image distance ($u$ and $v$) in terms of refractive index $n$ of the medium and $R$.

Hint:

Always define the variables and state the Cartesian sign convention explicitly when deriving optical formulas in board exams.
Remember that the paraxial approximation is the critical assumption that logically permits substituting angles with their tangents.

Answer 1

Concept:
• Refraction at a single spherical surface forms the foundational basis of image formation by lenses.

• The fundamental equation relates object distance, image distance, and the refractive indices of the respective media.

• We utilize the paraxial approximation, meaning the rays make exceedingly small angles with the principal axis.

Step 1: Ray Diagram Formulation
Let a point object $O$ be placed on the principal axis in a rarer medium of refractive index $n_1$.
The light ray strikes a convex spherical refracting surface separating the rarer medium from a denser medium of refractive index $n_2$.

A paraxial ray from $O$ is incident at point $A$ on the surface and bends towards the normal $CN$, forming a real image at $I$.

Step 2: Geometrical Analysis
Let the incident ray make an angle $\alpha$ with the principal axis, the refracted ray make an angle $\beta$, and the normal make an angle $\gamma$.
From the basic geometry of the triangles formed, the exterior angle is always equal to the sum of interior opposite angles.
In triangle $OAC$, the angle of incidence $i$ is given by $i = \alpha + \gamma$.
In triangle $AIC$, the angle of refraction $r$ is given by $\gamma = r + \beta$, which implies $r = \gamma - \beta$.

Step 3: Applying Paraxial Approximations
For paraxial rays, the point $A$ is assumed very close to the pole $P$, making the angles extremely small.
We can approximate the angles using their tangents:
$\alpha \approx \tan \alpha \approx \frac{AM}{PO} \approx \frac{AM}{-u}$.
$\beta \approx \tan \beta \approx \frac{AM}{PI} \approx \frac{AM}{v}$.
$\gamma \approx \tan \gamma \approx \frac{AM}{PC} \approx \frac{AM}{R}$.

Step 4: Snell's Law Application
According to Snell's law, for small angles, $n_1 i = n_2 r$.
Substituting the expressions for $i$ and $r$:
$n_1 (\alpha + \gamma) = n_2 (\gamma - \beta)$.
Substituting the tangent approximations alongside the Cartesian sign convention gives:
$n_1 \left(\frac{AM}{-u} + \frac{AM}{R}\right) = n_2 \left(\frac{AM}{R} - \frac{AM}{v}\right)$.
Canceling out the common altitude $AM$ from both sides leaves us with:
$\frac{-n_1}{u} + \frac{n_1}{R} = \frac{n_2}{R} - \frac{n_2}{v}$.
Rearranging the terms yields the final comprehensive relation:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
If the first medium is vacuum or air ($n_1 = 1$) and the second medium has a refractive index $n$ ($n_2 = n$), the formula simplifies to:
$\frac{n}{v} - \frac{1}{u} = \frac{n - 1}{R}$.